Problem 29 - The Calculus with Analytic Geometry

Question

In Exercises 29 through 32, solve for \(x\) and use absolute value bars to write the answer. \(\frac{x-a}{x+a}>0\)

Step-by-Step Solution

Answer

|x| > a

Step 1: Identify Critical Points

Set the numerator and the denominator equal to zero to find the critical points. The numerator is zero when \(x - a = 0\), thus \(x = a\). The denominator is zero when \(x + a = 0\), thus \(x = -a\).

Step 2: Determine Intervals

The critical points divide the real number line into three intervals: \((-\infty, -a)\), \((-a, a)\), and \((a, \infty)\).

Step 3: Test Each Interval

Select a test point from each interval and plug it into the inequality \(\frac{x-a}{x+a} > 0\) to determine if the inequality is satisfied. For \(x < -a\): Choose \(x = -a-1\). \(\frac{(-a-1)-a}{(-a-1)+a} = \frac{-2a-1}{-1} > 0\) is true. For \(-a < x < a\): Choose \(x = 0\). \(\frac{0-a}{0+a} = \frac{-a}{a} < 0\) is false. For \(x > a\): Choose \(x = a+1\). \(\frac{(a+1)-a}{(a+1)+a} = \frac{1}{2a+1} > 0\) is true.

Step 4: Write the Solution in Interval Form

Combine the intervals where the inequality is true. The solution is \(x < -a\) or \(x > a\). In interval notation, this is \((-\infty, -a) \cup (a, \infty)\).

Step 5: Use Absolute Value Bars

Express the solution using absolute value bars. Since \(x\) is outside the interval between \(-a\) and \(a\), the condition \(|x| > a\) captures this requirement.

Key Concepts

Critical Points

Critical points are values that divide the number line into different intervals where the sign of an expression may change. They are found by setting the numerator and denominator of a rational expression to zero. In solving the inequality \(\frac{x-a}{x+a}>0\), we set \((x-a)=0\) and \(x=-a\). This gives us the critical points \(x=a\) and \(x=-a\). These points are crucial because they tell us where the expression can potentially change its sign. Remember, always identify and mark these points before proceeding with further analysis.

Interval Notation

Interval notation is a method used to describe the set of solutions to inequalities. After finding critical points, we divide the real number line into intervals. For example, if our critical points are \(x = a\) and \(x = -a\), the intervals are \((-\infty, -a)\), \((-a, a)\), and \((a, \infty)\). Next, we test a point within each interval to determine where the inequality holds true. Finally, we combine the intervals where the inequality is valid. For the inequality \(\frac{x-a}{x+a} > 0\), the solution intervals are \((-\infty, -a)\) and \((a, \infty)\). In interval notation, this is written as \((-\infty, -a) \cup (a, \infty)\).

Absolute Value Inequalities

Absolute value inequalities express the distance of a number from zero on a number line. For example, the inequality \(|x| > a\) means x is more than 'a units' away from zero. To solve inequalities like \(\frac{x-a}{x+a}>0\), we often translate the final interval solution into an absolute value inequality. Given the solution intervals \((-\infty, -a) \cup (a, \infty)\), we can express this condition using absolute value as \(|x| > a\). This simply indicates that x lies outside the range between \(-a\) and \(a\). Using absolute value helps to condense the solution and makes it easier to understand and communicate.